$If{\cos }^4\theta {\sec }^2\alpha ,1/2and{\sin }^4\theta .\cos e{c}^2\alpha areinA.P.,then{\cos }^8\theta .{\sec }^6\alpha ,1/2and{\sin }^8\theta .\cos e{c}^6\alpha arein$

$\begin{array}{l}{\cos }^4\theta .{\sec }^2\alpha ,1/2,{\sin }^4\theta ,cose{c}^2\alpha areinA.P.\Rightarrow \frac{{\cos }^4\theta }{{\cos }^2\alpha }+\frac{{\sin }^4\theta }{{\sin }^2\alpha }=2.\frac{1}{2}=1\\ \Rightarrow {\cos }^4\theta {\sin }^2\alpha +{\sin }^4\theta {\cos }^2\alpha ={\sin }^2\alpha {\cos }^2\alpha \\ \Rightarrow (1-{\sin }^2\theta ){\cos }^2\theta {\sin }^2\alpha +{\sin }^4\theta (1-{\sin }^2\alpha )={\sin }^2\alpha (1-{\sin }^2\alpha )\\ \Rightarrow {\cos }^2\theta {\sin }^2\alpha +{\sin }^4\theta -{\sin }^2\theta {\sin }^2\alpha ({\cos }^2\theta +{\sin }^2\theta )={\sin }^2\alpha -{\sin }^4\alpha \\ \Rightarrow {\sin }^4\theta +{\sin }^4\alpha -{\sin }^2\theta {\sin }^2\alpha -{\sin }^2\alpha (1-{\cos }^2\theta )=0\Rightarrow {\sin }^4\theta +{\sin }^4\alpha -2{\sin }^2\theta {\sin }^2\alpha =0\\ \Rightarrow {({\sin }^2\theta -{\sin }^2\alpha )}^2=0\Rightarrow {\sin }^2\theta {\sin }^2\alpha \Rightarrow {\cos }^2\theta ={\cos }^2\alpha \\ \therefore {\cos }^8\theta {\sec }^6\alpha +{\sin }^8\theta \cos e{c}^6\alpha ={\cos }^8\theta .{\sec }^6\theta +{\sin }^8\theta .\cos e{c}^2\theta ={\cos }^2\theta +{\sin }^2\theta =1.\\ \therefore {\cos }^8\theta .{\sec }^6\alpha ,1/2,{\sin }^8\theta .\cos e{c}^6\alpha areinA.P.\end{array}$